I'm not really sure if this problem belongs in this forum. It is from Tipler chapter 3 on Quantization of Charge, Light, and Energy.

Problem 4: A cosmic ray proton approaches Earth vertically at the equator, where the horizontal component of Earth's magnetic field is 3.5 X 10^-5 T. If the proton is moving at 3.0X10^6 m/s, what is the ratio of the magnetic force to the gravitational force on the proton?

My answer is Fb/Fg = BQv sin theta / mg and since the velocity is perpendicular to the magnetic force then sin theta = 1.